Q. A particle of mass 'm' is projected with a velocity $v = kV_e$ (k < 1) from the surface of the earth. The maximum height above the surface reached by the particle is:
A) $R\left(\frac{k}{1+k}\right)^2$
B) $\frac{R^2k}{1+k}$
C) $\frac{Rk^2}{1-k^2}$
D) $R\left(\frac{k}{1-k}\right)^2$
Explanation
[Solution] $\frac{1}{2}mk^2\frac{2GM}{R} - \frac{GMm}{R} = -\frac{GMm}{R+h}$ gives $h = \frac{Rk^2}{1-k^2}$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.